SOLUTION: Two numbers differ by 8 and their name product is 209. Find the numbers

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Question 1194290: Two numbers differ by 8 and their name product is 209. Find the numbers
Found 3 solutions by Boreal, ikleyn, greenestamps:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
x and y
I assume their product is 209
xy=209
x-y=8
x=y+8
so (y+8)*y=209
y^2+8y-209=0
(y+19)(y-11)=209
The numbers are 11 and 19 or -11 and -19
Answer by ikleyn(52771)   (Show Source): You can put this solution on YOUR website!
.

Let x be the number exactly half way between our two numbers.


Then the two numbers are x+4 and x-4;  their product is (x+4)*(x-4) = x^2 - 16.


So we have this equation for x

    x^2 - 16 = 209

    x^2 = 209 + 16 = 225.


Thus  x =  = +/- 15/


The problem has two solutions.


One solution is the pair  (15-4,15+4) = (11,19).


Another solution is the pair  (-15-4,-15+4) = (-19,-11).

Solved.



Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


Notice that, in the formal algebraic solution from @boreal, you have to factor a quadratic expression by finding two numbers whose difference is 8 and whose product is 209.

But that is exactly what the original problem asks you to do -- so that formal algebraic method doesn't help you get any closer to the answer.

If you need a formal algebraic solution, use the nice "trick" shown by tutor @ikleyn. As she says in her response, it is a very useful method applicable to many problems.

But if formal algebra is not required, clearly the fastest way to the answer is to find how the product 209 factors.

209 = 11*19

So the two numbers are 11 and 19.

Then, since the problem allows negative answers, another solution is -11 and -19.
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